Benzoate ion hydrolyzes in water to form a solution that is not neutral.
If we call the benzoate ion Bz^-, then
.........Bz^- + HOH ==> HBz + OH^-
I........1.0..............0.....0
C........-x..............x.....x
E.......1-x..............x......x

Kb for Bz^- = (Kw/Ka for HBz) = x*x/(1-x) and solve for x = (OH^-),then convert to pH.
HBz is benzoic acid. You can find Ka for HBz in you text/notes I'm sure but it's approx 10^-5.